Advertisements
Advertisements
प्रश्न
Find the value of 'a' and 'b' if:
`(sqrt243)^"a" ÷ 3^("b" + 1)` = 1 and `27^"b" - 81^(4 -"a"/2)` = 0
Advertisements
उत्तर
`(sqrt243)^"a" ÷ 3^("b" + 1)` = 1 and `27^"b" - 81^(4 -"a"/2)` = 0
⇒ `(sqrt(3^5))^"a" ÷ 3^("b" + 1) and (3^3)^"b" - (3^4)^(4 - "a"/2)` = 0
⇒ `(3^5)^("a"/2) ÷ 3^("b" + 1) = 1 and 3^(3"b") - (3^4)^(4 - "a"/2)` = 0
⇒ `3^(((5"a")/2)) ÷ 3^("b" + 1) = 1 and 3^((3"b")) - 3^(4(4 - "a"/2)` = 0
⇒ `3^(((5"a")/2 - "b" - 1)) = 1 and 3^((3"b")) - 3^(16 - 2"a")` = 0
⇒ `3^(((5"a")/2 - "b" - 1)) = 3^° and 3^(3"b") = 3^(16 - 2"a")`
⇒ `(5"a")/(2) - "b" - 1 = 0 and 3"b"` = 16 - 2a
⇒ `(5"a")/(2) - "b" = 1 and 2"a" + 3"b"` = 16
⇒ 5a - 2b = 2 and 2a + 3b = 16
Multiply the equations by 3 and 2 respectively.
⇒ 15a - 6b = 6 and 4a + 6b = 32
Adding the equations,
19a = 38
⇒ a = 2
Substitute the value of ain 5a - 2b = 2 to find b.
5a - 2b = 2
⇒ 5(2) - 2b = 2
⇒ 10 - 2b = 2
⇒ b = 4
Hence, a = 2 and b = 4.
APPEARS IN
संबंधित प्रश्न
Solve for x:
`3^(4x + 1) = (27)^(x + 1)`
Solve for x : 3(2x + 1) - 2x + 2 + 5 = 0
Find the values of m and n if :
`4^(2m) = ( root(3)(16))^(-6/n) = (sqrt8)^2`
Evaluate the following:
`(8/27)^((-2)/3) - (1/3)^-2 - 7^0`
Solve for x:
22x+1= 8
Solve for x:
`9 xx 3^x = (27)^(2x - 5)`
Solve for x:
`sqrt((3/5)^(x + 3)) = (27^-1)/(125^-1)`
Find the value of k in each of the following:
`(sqrt(9))^-7 xx (sqrt(3))^-5` = 3k
Find the value of 'a' and 'b' if:
92a = `(root(3)(81))^(-6/"b") = (sqrt(27))^2`
Prove the following:
`("a"^"m"/"a"^"n")^("m"+"n"+1) ·("a"^"n"/"a"^1)^("n" + 1-"m").("a"^1/"a"^"m")^(1+"m"-"n")`
